Question

If $a^2-12ab+{4b}^2=0,$ prove that $\log (a+2b)=\frac{1}{2}(\log a+\log b)+2\log 2$

Answer 1

We have

$a^2-12ab+4b^2=0$

$\Rightarrow {\left(a+2b\right)}^2-16ab=0,\Rightarrow {\left(a+2b\right)}^2=16ab...\left(1\right)$

$Taking\log onbothsidesineq....\left(1\right)$

$2\log \left(a+2b\right)=\log \left(16ab\right)$

$\log \left(a+2b\right)=\frac{1}{2}\log \left(16ab\right)$

$\log \left(a+2b\right)=\frac{1}{2}\left(\log a+\log b\right)+\frac{1}{2}\left(\log 16\right)$

$\log \left(a+2b\right)=\frac{1}{2}\left(\log a+\log b\right)+\log {16}^{\frac{1}{2}}$

$\log \left(a+2b\right)=\frac{1}{2}\left(\log a+\log b\right)+\log 4$

$\log \left(a+2b\right)=\frac{1}{2}\left(\log a+\log b\right)+2\log 2$